# Full text of "Treatise On Analysis Vol-Ii"

## See other formats

```350       XV   NORMED ALGEBRAS AND SPECTRAL THEORY

then it is clear thatfxo is a linear form on A, and we have

f^s**) = (U(s*)U(s) - x0 | jc0) = (J7(s) • x0 | U(s) • *0) =
by virtue of (15.5.1). The corresponding Hilbert form is

(15.6.5)                        gxo(s, t) = (ii(s) • x0 | u(i) - x0).

For example, if A = J^(H) where H is a Hilbert space of finite dimension n,
and 1A : T\-+T is the identity representation of A in H, then the form/xo is

calculated explicitly as follows : if (e^ ^ t^n is an orthonormal basis of H, and

n

(f f;) the matrix of T with respect to this basis, then for x0 = Y l,ef we have

i=l

(15.6.6)                           /JTO-EWv-

*»J

When studying positive linear forms of the type (15.6.4) we may always
assume that x0 is a totalizer for U (15.5), because/^ is unchanged when we
replace U by its restriction to the stable subspace of H which is the closure of
the stable subspace generated by x0 and the U(s) • x0 (s e A). Under this
additional assumption, the form fXQ determines the representation U up to
equivalence :

(15.6.7)    Let U, U' be two representations of A in Hilbert spaces H, H',
respectively, and suppose that U (resp. U') has a totalizer x0 (resp. x'0). Then if
(U(s) * x0 1 XQ) = (U'(s) - XQ | x'o)for all s e A, the representations U and U' are
equivalent.

For all s, tin A we have
(15.6.7.1)

Since the vectors U(t) • x0 (resp. Uf(t)-xf0) form a dense subspace H0
(resp. HO) of H (resp. H'), this proves already that U(s) - x0 = 0 if and
only if U'(s) 'x'0 = Q. It follows that, for each z e H0 and each s e A such
that U(s)-x0 = z, the vector U'(s)-xr0 is constant and equal to say
z' = T- zeHJ). By (15.6.7.1) the mapping Tis an isomorphism of the pre-
hilbert space H0 onto the prehilbert space H'0 , which extends uniquely to an
isomorphism (also denoted by 7") of the Hilbert space H onto the Hilbert
space H' (by virtue of (5.5.4) and the principle of extension of identities). Ite topologi-
```